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10.1.4 Potential difference and voltmeters

Potential difference

Definition

Potential difference

Potential difference is the energy transferred per unit of charge as charge moves between two points in a circuit.

Definition

Volt

One volt is one joule of energy transferred per coulomb of charge, so 1 V=1 J C−11\ \text{V}=1\ \text{J C}^{-1}1 V=1 J C−1.

  1. The potential difference between two points is the energy transferred per unit charge as charge moves from one of those points to the other.
  2. A cell does work on the charge that passes through it, so the charge leaves the cell carrying energy that it did not have before.
  3. As that charge passes through a component it transfers the energy to the component, which then fills a store or passes the energy on to the surroundings.
  4. The potential difference across the supply is the energy the supply gives to each unit of charge that passes through it.
  5. The potential difference across a component is the energy transferred by each unit of charge as it passes through that one component.
  6. The unit is the volt, and a potential difference of 1 V1\ \text{V}1 V means that 1 J1\ \text{J}1 J of energy is transferred for each 1 C1\ \text{C}1 C of charge, so 1 V=1 J C−11\ \text{V}=1\ \text{J C}^{-1}1 V=1 J C−1.
  7. The quantity is given the symbol VVV and its unit is also written V\text{V}V, so a reading written as V=3.0 VV=3.0\ \text{V}V=3.0 V is a potential difference of three volts.
  8. A larger potential difference means each unit of charge is given more energy, and that is what makes the supply push the charge round the circuit more strongly.
  9. Two cells joined in series, both facing the same way, give twice the potential difference of one cell, so each unit of charge is given twice the energy.
  10. A potential difference is always a difference between two points, which is why it is measured across a component and never at a single point.
Key Idea

A potential difference of 1 V1\ \text{V}1 V transfers 1 J1\ \text{J}1 J of energy for every 1 C1\ \text{C}1 C of charge that passes, which is why the volt is the same thing as 1 J C−11\ \text{J C}^{-1}1 J C−1.

Using a voltmeter

Definition

Voltmeter

A voltmeter is a meter connected in parallel with a component to measure the potential difference across it.

  1. A voltmeter measures the potential difference between two points, so its two terminals are connected to the two ends of the component being tested.
  2. That connection places the voltmeter in parallel with the component, on a branch of its own alongside it.
  3. It has to be connected this way because a difference can only be found by touching both of the points it lies between, and only a branch across the component reaches both of those points.
  4. A voltmeter is built with a very high resistance, so almost no charge is diverted along its branch and the circuit it is measuring is barely disturbed.
  5. A voltmeter is never connected in series, because in the loop it would no longer sit across a component, and its very high resistance would choke the flow of charge in the whole circuit.
  6. A digital voltmeter gives the reading directly, and its resolution is the smallest change it can display, which is 0.01 V0.01\ \text{V}0.01 V on a meter that shows two decimal places.
  7. A reading is quoted to the precision the meter allows, so a meter with a resolution of 0.01 V0.01\ \text{V}0.01 V gives 2.35 V2.35\ \text{V}2.35 V and not 2.3 V2.3\ \text{V}2.3 V or 2.352 V2.352\ \text{V}2.352 V.
  8. An analogue voltmeter is read with the eye directly in front of the pointer, so that a slanted line of sight does not shift the value read off the scale.
  9. A meter that does not read 0 V0\ \text{V}0 V when it is disconnected has a zero error, and every reading it gives is shifted by that same amount until the meter is adjusted or the error is subtracted.
Common Mistake
  • Do not connect a voltmeter in series with the component. It belongs on a branch across the component.
  • Do not describe potential difference as something that flows round a circuit. Charge flows, and potential difference is a difference between two points.
  • Do not confuse the symbol VVV for the quantity with the unit symbol V\text{V}V for the volt.
  • Do not quote a reading beyond the resolution of the meter, such as writing 2.345 V2.345\ \text{V}2.345 V from a display that shows only two decimal places.
  • Do not leave a zero error uncorrected and then call the readings accurate.

Sharing in series

  1. In a series circuit the charge passes through every component in turn, so the energy each unit of charge received from the supply is transferred in stages, part of it in each component.
  2. The potential differences across the components therefore add up to the potential difference across the supply: Vsupply=V1+V2+…V_{\text{supply}}=V_1+V_2+\dotsVsupply​=V1​+V2​+…
  3. The reason lies in energy per unit charge: the total energy one unit of charge transfers on its way round the loop is the sum of the amounts it transfers in each component, and that total must equal the energy the supply gave it.
  4. If the readings did not add up, charge would arrive back at the supply carrying more or less energy than it set out with, and energy would not be conserved.
  5. Adding a second identical lamp to a series loop therefore halves the potential difference across the first lamp, because the same total is now shared between two lamps.
  6. Each component can be measured separately by moving one voltmeter from one component to the next, and the readings can then be added and checked against a reading taken across the supply.
  7. A voltmeter connected across two components in series reads the sum of their two separate potential differences, because it spans both of them.
  8. A voltmeter placed across a plain connecting wire reads almost 0 V0\ \text{V}0 V, because hardly any energy is transferred per unit of charge in the wire itself.
Example

Sharing between two components

  • A supply of 6.0 V6.0\ \text{V}6.0 V is connected in series with a lamp and a resistor, and a voltmeter across the lamp reads 2.5 V2.5\ \text{V}2.5 V.
  • In series the potential differences add, so Vsupply=Vlamp+VresistorV_{\text{supply}}=V_{\text{lamp}}+V_{\text{resistor}}Vsupply​=Vlamp​+Vresistor​.
  • Rearranging for the unknown gives Vresistor=Vsupply−VlampV_{\text{resistor}}=V_{\text{supply}}-V_{\text{lamp}}Vresistor​=Vsupply​−Vlamp​.
  • Substituting the readings gives Vresistor=6.0−2.5V_{\text{resistor}}=6.0-2.5Vresistor​=6.0−2.5.
  • The potential difference across the resistor is 3.5 V3.5\ \text{V}3.5 V.
  • As a check, 2.5+3.5=6.0 V2.5+3.5=6.0\ \text{V}2.5+3.5=6.0 V, which matches the supply reading.

Parallel branches

  1. In a parallel circuit every branch is connected between the same two points, and when the branches sit directly across the supply those two points are the supply terminals themselves.
  2. Charge taking any one branch therefore travels between the same pair of points, so it transfers the same energy per unit charge whichever branch it takes.
  3. The potential difference across every parallel branch is the same, and it equals the potential difference across the supply when the branches are connected directly across it.
  4. This is why identical lamps in parallel all glow equally brightly, and why each of them glows as brightly as a single lamp would on its own across that supply.
  5. Adding another branch leaves the potential difference across the existing branches unchanged, because the two points they are joined between have not moved.
  6. Switching off one branch also leaves the potential difference across the others unchanged, which is why the remaining lamps do not dim.
  7. Within a single branch that holds two components, those two components are in series with each other, so their potential differences share the branch total between them.
Example

Branches across a supply

  • A 12 V12\ \text{V}12 V supply feeds two branches, one holding a motor alone and the other holding a lamp in series with a resistor.
  • Both branches are joined directly across the supply, so the potential difference across each whole branch is 12 V12\ \text{V}12 V.
  • The motor is alone on its branch, so the potential difference across the motor is V=12 VV=12\ \text{V}V=12 V.
  • On the other branch the two components are in series, so Vlamp+Vresistor=12 VV_{\text{lamp}}+V_{\text{resistor}}=12\ \text{V}Vlamp​+Vresistor​=12 V.
  • A voltmeter across the resistor reads 8.0 V8.0\ \text{V}8.0 V, so rearranging gives Vlamp=12−8.0V_{\text{lamp}}=12-8.0Vlamp​=12−8.0.
  • The potential difference across the lamp is 4.0 V4.0\ \text{V}4.0 V.

Unequal components

  1. Two components in series only take equal shares of the supply potential difference when they are identical.
  2. The component that opposes the flow of charge more strongly takes the larger share, because more energy is transferred by each unit of charge in passing through it.
  3. The component that opposes the flow less takes the smaller share, and the two shares still add to the supply potential difference.
  4. A lamp in series with a very small resistor therefore carries almost all of the supply potential difference, and the reading across the resistor is only a fraction of a volt.
  5. Replacing one of a series pair with a component that opposes the flow more strongly raises the reading across it and lowers the reading across its partner by exactly the same amount, since the total is fixed by the supply.
  6. Comparing two voltmeter readings taken across a series pair is therefore a direct way of comparing how strongly the two components oppose the flow of charge.
Exam technique

Working with potential difference

  • Quote the rule before substituting, writing Vsupply=V1+V2V_{\text{supply}}=V_1+V_2Vsupply​=V1​+V2​ for a series loop.
  • Name the two points a reading belongs to, so write the potential difference across the lamp rather than the potential difference at the lamp.
  • Convert a reading given in millivolts before adding it to another, so 250 mV250\ \text{mV}250 mV becomes 0.250 V0.250\ \text{V}0.250 V.
  • For a parallel arrangement, state that the branches are joined between the same two points and that this is why their potential differences are equal.
  • When describing a measurement, say that the voltmeter is connected in parallel with the named component, and give the unit as V\text{V}V.
Self review
  • Define potential difference and state what 1 V1\ \text{V}1 V means in joules and coulombs.
  • Explain why a voltmeter must be connected in parallel with the component being tested.
  • State the rule linking the potential differences across components in series to that of the supply, and give the energy reason for it.
  • State the rule for the potential differences across the branches of a parallel circuit.
  • Explain which of two unequal components in series takes the larger share of the supply potential difference.
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10.1.4 Potential difference and voltmeters Revision Guide

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Revision notes for Edexcel GCSE Physics 10.1.4 Potential difference and voltmeters: explanations and worked examples.

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